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Title: Empirical Formula and Molecular Formula
Description: This material explains, in detail, the procedure for determining the empirical and molecular formulas of compounds in stepwise manner.
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Study Guide

Empirical and Molecular Formulas

Empirical Formula
The empirical formula of a compound shows the relative number of atoms of each element in the
compound
...
Hence, the subscript is actually the ratio of
the number of moles of the elements in the compound
...

Molecular formulas show that actual number of moles of each element in a compound
...

The molecular formula for Nitrogen dioxide is
NO2
This is the same as its empirical formula
...

If we know the empirical formula and the molecular weight of the compound, we can easily
determine the whole number multiple to which we should multiply the empirical formula to get the
molecular formula
...
From percent composition data
...

1
...
Remember that percent composition is
based on mass ratios
...

2
...

3
...
This will give us the mole ratio for each element
...
Round the mole ratios to the nearest whole number and write the empirical
formula using the mole ratio of each element as their respective subscript
...
Dela Cruz

Study Guide

Empirical and Molecular Formulas

Example 1
...
5% Silver, 8
...
3% Oxygen?
Step 1
...
Given that the sample has a mass of
100g, then it is composed of 63
...
2g of Nitrogen and 28
...


Step 2
...

1 π‘šπ‘œπ‘™ 𝐴𝑔

(63
...
87𝑔 )=
...
2g N) ( 14
...
586 mol N
1 π‘šπ‘œπ‘™ 𝑂

(28
...
00𝑔 )= 1
...
Divide the amount in moles of each element by the smallest number of moles
calculated
...
586 mol
...

For silver

...
586 π‘šπ‘œπ‘™

For Nitrogen
For oxygen


...
586 π‘šπ‘œπ‘™
1
...
586 π‘šπ‘œπ‘™

= 1
...
00
= 3
...
Round the mole ratios to the nearest whole number and write the empirical formula
using the mole ratio of each element as their respective subscript
...

Example 2
...
8% Fe,
21
...
1% O
...
Assume that the sample has a mass of 100g
...
8g of the mass is contributed by Iron, 21
...
1g by Oxygen
...
Convert the mass of each element to mole using the molar mass of the element
...
8g Fe) (

1 π‘šπ‘œπ‘™ 𝐹𝑒
55
...
1g S) (

1 π‘šπ‘œπ‘™ 𝑆
32
...
659 mol Fe

)=
...
1g O) ( 16
...
63 mol O
2

Reymon T
...
Divide the amount in moles of each element by the smallest number of moles
calculated
...
658 mol
...

For Iron

...
00


...
658 π‘šπ‘œπ‘™

For Sulfur

= 1
...
658 π‘šπ‘œπ‘™
2
...
99


...
Round the mole ratios to the nearest whole number and write the empirical formula
using the mole ratio of each element as their respective subscript
...

Example 3
...
8% C and 17
...
What is the empirical
formula of the hydrocarbon?

Step 1
...
If we assume that the sample is 100g,
then the mass of carbon in the substance is 82
...
2g
...
Convert the mass of each element to mole using the molar mass of the element
...
8g C) ( 12
...
9 mol C
(17
...
00𝑔

)= 17
...
Divide the amount in moles of each element by the smallest number of moles
calculated
...
9 mol
...

For Carbon
For Hydrogen

6
...
00

6
...
2 π‘šπ‘œπ‘™
6
...
49

3

Reymon T
...
Round the mole ratios to the nearest whole number and write the empirical formula
using the mole ratio of each element as their respective subscript
...
49 β‰ˆ 2
...
So, we
multiply both ratios by 2 to get whole number ratios for our empirical formula
...
9 π‘šπ‘œπ‘™
6
...
2 π‘šπ‘œπ‘™
6
...
00 X 2 = 2
= 2
...

B
...

For problems in which the mass of the sample’s component elements is given, the following
are the steps we need to do to determine the empirical formula of the substance
...
Convert the mass of each element to mole using the molar mass of the element
...
Divide the amount in moles of each element by the smallest number of moles
calculated
...

3
...

Example 1
...
56 g of Hydrogen, 12
...
What is the empirical formula of ethanol?
Step 1
...

To determine the mass of carbon, we simply subtract the masses of Hydrogen and Oxygen from
the mass of the sample
...
56g-12
...
26g
...
26g C) ( 12
...
52mol C
(12
...
56g H) (

1 π‘šπ‘œπ‘™ 0
16
...
00𝑔

)=
...
56 mol H

Step 2
...

We can see from the results of step 2 that the smallest amount in moles is
...
We divide all
molar amounts with this number to get the mole ratio for each element
...
52 π‘šπ‘œπ‘™

...
761 π‘šπ‘œπ‘™

...
56 π‘šπ‘œπ‘™

...
99
= 1
...
99

Step 3
...

Mole ratio C = 2
4

Reymon T
...

Example 2
...
4g of Calcium, 77
...
86g of Hydrogen
...
Convert the mass of each element to mole using the molar mass of the element
...
4g Ca) (

1 π‘šπ‘œπ‘™ πΆπ‘Ž
40
...
43 mol Ca

1 π‘šπ‘œπ‘™ 0

(77
...
00𝑔 )= 4
...
86g H) (

1 π‘šπ‘œπ‘™ 𝐻
1
...
86 mol H

Step 2
...

We can see from the results of step 2 that the smallest amount in moles is 2
...
We divide all
molar amounts with this number to get the mole ratio for each element
...
43 π‘šπ‘œπ‘™
2
...
86 π‘šπ‘œπ‘™
2
...
86 π‘šπ‘œπ‘™
2
...
00
= 2
...
00

Step 3
...

Mole ratio Ca = 1
Mole ratio O = 2
Mole ratio H = 2
The empirical formula is CaO2H2 or Ca(OH)2
...
A halogenated Hydrocarbon is composed of 18
...
08g Hydrogen and
53
...
What is the empirical mass of the substance?
Step 1
...

1 π‘šπ‘œπ‘™ 𝐢

(18
...
01𝑔 )= 1
...
08g H) (

1 π‘šπ‘œπ‘™ 𝐻
1
...
08 mol H

1 π‘šπ‘œπ‘™ 𝐢𝑙

(53
...
45𝑔 )= 1
...
Divide the amount in moles of each element by the smallest number of moles
calculated
...
51 mol
...

5

Reymon T
...
53 π‘šπ‘œπ‘™

For Carbon

1
...
08 π‘šπ‘œπ‘™

For Hydrogen

1
...
51 π‘šπ‘œπ‘™

For Chlorine

1
...
01
= 2
...
00

Step 3
...

Mole ratio C = 1
Mole ratio H = 2
Mole ratio Cl = 1
The empirical formula is CH2Cl
...
We do this by getting the ratio of the molecular
molar mass and the empirical molar mass (mass based on the empirical formula)
...
The molar mass of a substance is 42 g/mol
...
Calculate the empirical mass of the substance
...

Element in the
compound

Molar mass of the
element

Carbon
Hydrogen

12
...
00g/mol
Molar mass of CH2

Subscript

1
2

Total mass
contributed by the
element
12
...
00g/mol
14
...
Divide the molar mass of the substance with its empirical mass to get the whole
number multiple
...
01𝑔/π‘šπ‘œπ‘™
Whole number multiple = 3
Step 3
...

3(CH2)
C3H6
The molecular formula of the substance is C3H6
...
Dela Cruz

Study Guide

Empirical and Molecular Formulas

Example 2
...
9% Hydrogen, and
27
...
What is the molecular formula of the substance if its molar mass is 118g/mol?
Step 1
...

a
...
This means that 61g of the sample is
Carbon, 11
...
1g is Oxygen
...
We then convert all given masses to moles using the molar mass of each element
...
9g H) (

For Oxygen

(27
...
08 mol C

12
...
00𝑔
1π‘šπ‘œπ‘™ 𝑂
16
...
9 mol H
) = 1
...
We divide the molar amounts by the smallest mole calculated
...
69
...

5
...
69 π‘šπ‘œπ‘™
11
...
69 π‘šπ‘œπ‘™
1
...
69 π‘šπ‘œπ‘™

= 3
...
04
= 1
...
Round the mole ratios to the nearest whole number and write the empirical formula
...
We need to determine the empirical molar mass of the substance
...
01g/mol
3
36
...
00g/mol
7
7
...
00g/mol
1
16
...
03g/mol
Step 4
...

Whole number multiple =

π‘šπ‘œπ‘™π‘’π‘π‘’π‘™π‘Žπ‘Ÿ π‘šπ‘œπ‘™π‘Žπ‘Ÿ π‘šπ‘Žπ‘ π‘ 
π‘’π‘šπ‘π‘–π‘Ÿπ‘–π‘π‘Žπ‘™ π‘šπ‘œπ‘™π‘Žπ‘Ÿ π‘šπ‘Žπ‘ π‘ 
118𝑔/π‘šπ‘œπ‘™

Whole number multiple = 59
...
Multiply the subscript of each element in the empirical formula by the whole number ratio
...

7

Reymon T
Title: Empirical Formula and Molecular Formula
Description: This material explains, in detail, the procedure for determining the empirical and molecular formulas of compounds in stepwise manner.
Buy These NotesPreview